You have found the following ages (in years) of 6 turtles. Those turtles were randomly selected from the 28 turtles at your local zoo: $ 4,\enspace 50,\enspace 63,\enspace 2,\enspace 82,\enspace 99$ Based on your sample, what is the average age of the turtles? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 28 turtles, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{4 + 50 + 63 + 2 + 82 + 99}{{6}} = {50\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {2116} + {0} + {169} + {2304} + {1024} + {2401}} {{6 - 1}} $ {s^2} = \dfrac{{8014}}{{5}} = {1602.8\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{1602.8\text{ years}^2}} = {40\text{ years}} $ We can estimate that the average turtle at the zoo is 50 years old. There is also a standard deviation of 40 years.